Proper use of comparison theorum?

[PCSL]

\int_{0}^{\frac{pi}{2}} \frac{cos(x)}{x^{1/2}} dx

\int_{0}^{\frac{pi}{2}} \frac{cos(x)}{x^{1/2}} dx \le \int_{0}^{\frac{pi}{2}} \frac{1}{x^{1/2}} dx

This would mean that the original equation was convergent. Was my reasoning correct in making the numerator 1 since the max value of cos(x) is 1?

Also, are there any hints you guys can give me for using the comparison theorem? Thanks.

 

[LCKurtz]

\int_{0}^{\frac{pi}{2}} \frac{cos(x)}{x^{1/2}} dx

\int_{0}^{\frac{pi}{2}} \frac{cos(x)}{x^{1/2}} dx \le \int_{0}^{\frac{pi}{2}} \frac{1}{x^{1/2}} dx

This would mean that the original equation was convergent. Was my reasoning correct in making the numerator 1 since the max value of cos(x) is 1?

Also, are there any hints you guys can give me for using the comparison theorem? Thanks.

Yes. It looks like you have the idea. Generally if you think something converges you would overestimate the numerator and/or underestimate the denominator to get an integrand that is both

1. Easier
2. Convergent.

And everything is just the opposite if you are trying to show an integral diverges.

 

[PCSL]

Thanks!

Would this one be correct?

\int_{\frac{1}{2}}^{1} \frac{1}{x(1-x)^{\frac{1}{3}}} dx \le \int_{\frac{1}{2}}^{1} \frac{1}{(1-x)^{\frac{1}{3}}} dx

Also, I'm a little confused on how I would break up

\int_{0}^{\frac{1}{3}} \frac{sec(\pi x)}{1-3x} dx

Could I have a hint?

 

[Dick]

1/(x*(1-x)^(1/3)) is GREATER than 1/((1-x)^(1/3)) on [1/2,1). Do you see why? (1/x) is GREATER than 1. For the second one, sketch a graph. What are the max and min values of sec(pi*x) on [0,1/3]?

 

[PCSL]

1/(x*(1-x)^(1/3)) is GREATER than 1/((1-x)^(1/3)) on [1/2,1). Do you see why? (1/x) is GREATER than 1. For the second one, sketch a graph. What are the max and min values of sec(pi*x) on [0,1/3]?

Haha, now the second one makes total sense but I'm lost on the first one (1/(x(1-x)^(1/3). I have to find an equation that is greater than the original correct?

 

[Dick]

Haha, now the second one makes total sense but I'm lost on the first one (1/(x(1-x)^(1/3).

I assume your given function is (1/(x(1-x)^(1/3)). Split it into (1/x)*(1/(1-x)^(1/3)). What are the max and min values of (1/x) for x in the range [1/2,1]?

 

[PCSL]

I assume your given function is (1/(x(1-x)^(1/3)). Split it into (1/x)*(1/(1-x)^(1/3)). What are the max and min values of (1/x) for x in the range [1/2,1]?

Could I do

\int_{\frac{1}{2}}^{1} \frac{1}{x(1-x)^{\frac{1}{3}}} dx \le \int_{\frac{1}{2}}^{1} \frac{1}{2(1-x)^{\frac{1}{3}}} dx

 

[Dick]

Could I do

\int_{\frac{1}{2}}^{1} \frac{1}{x(1-x)^{\frac{1}{3}}} dx \le \int_{\frac{1}{2}}^{1} \frac{1}{2(1-x)^{\frac{1}{3}}} dx

Close. 1/x <= 2. So (1/x)*(1/(1-x)^(1/3))<=2*(1/(1-x)^(1/3)). I hope that's what you meant.

 

[PCSL]

Close. 1/x <= 2. So (1/x)*(1/(1-x)^(1/3))<=2*(1/(1-x)^(1/3)). I hope that's what you meant.

Yes! Thank you for your help!